Unit 8 · Physical Chemistry

Quantitative Chemical Equilibrium

Kc, Kp, Kx, Kc-Kp relationship, ICE tables, reaction quotient Q, Le Chatelier, industrial applications.

8.1–8.3

Equilibrium Constant Kc

Dynamic EquilibriumA system reaches dynamic equilibrium when the rate of forward reaction = rate of reverse reaction. Concentrations remain constant (not necessarily equal). Achieved in closed systems.

The Equilibrium Constant Kc

For the general reaction: aA + bB ⇌ cC + dD

KἾ = [C]ἠ[D]ἡ / [A]ἐ[B]ἡ Rules: • Concentrations at equilibrium only (not initial) • Pure solids and pure liquids omitted (activity = 1) • KἾ is dimensionless (if using standard concentrations c° = 1 mol/L) • KἾ depends only on temperature (not concentration, pressure, or catalyst) Characteristics: KἾ >> 1: products strongly favoured (equilibrium lies right) KἾ << 1: reactants strongly favoured (equilibrium lies left) KἾ ≈ 1: comparable amounts of reactants and products Relationship between KἾ values: If reaction is reversed: KἾ’ = 1/KἾ If equation multiplied by n: KἾ’ = (KἾ)ⁿ If two equations added: KἾ(overall) = KἾ(1) × KἾ(2)
8.4–8.5

Kp and the Relationship Kp = Kc(RT)Δn

Kp Definition

For gas-phase reactions: Kρ expressed in terms of partial pressures pᵧ: Kρ = (pṢ)ἠ(pṤ)ἡ / (pṢ)ἐ(pṢ)ἡ where pᵧ in atm or Pa Relationship Kρ = KἾ(RT)∆ⁿ where Δn = (moles gaseous products) − (moles gaseous reactants) Examples: N₂ + 3H₂ ⇌ 2NH₃: Δn = 2−4 = −2. Kρ = KἾ(RT)⁻² → Kρ < KἾ H₂ + I₂ ⇌ 2HI: Δn = 2−2 = 0. Kρ = KἾ(RT)⁰ = KἾ (same!) PCl₅ ⇌ PCl₃ + Cl₂: Δn = 2−1 = +1. Kρ = KἾ(RT)¹ → Kρ > KἾ
8.6

Calculations on Kc and Kp

Example 1

ICE Table for Kc

H₂(g) + I₂(g) ⇌ 2HI(g). Initial: [H₂]=1.00, [I₂]=1.00, [HI]=0. At equilibrium [HI]=1.56 mol/L. Find KἾ.

1
ICE table: Change in HI = +1.56. Change in H₂ = −0.78; Change in I₂ = −0.78
2
Equilibrium: [H₂] = 1.00−0.78 = 0.22; [I₂] = 0.22; [HI] = 1.56
3
KἾ = [HI]² / ([H₂][I₂]) = (1.56)² / (0.22 × 0.22) = 2.434/0.0484 = 50.3
Example 2

Converting Kc to Kp

N₂O₄(g) ⇌ 2NO₂(g). KἾ = 4.63×10⁻³ at 25°C. Find Kρ. (R = 0.0821, T = 298 K)

1
Δn = 2 − 1 = +1
2
Kρ = KἾ(RT)∆ⁿ = 4.63×10⁻³ × (0.0821 × 298)¹ = 4.63×10⁻³ × 24.47 = 0.113
8.7

Reaction Quotient QἾ vs KἾ

Using Q to Predict Direction

Q is calculated using the same expression as K but with current concentrations (not equilibrium):

No videos added yet for this unit.

If Q < KἾ: reaction proceeds forward (→) to produce more products If Q > KἾ: reaction proceeds backward (←) to produce more reactants If Q = KἾ: system IS at equilibrium; no net change Le Chatelier's principle: • Increase [reactant]: Q < K → forward; products increase • Increase [product]: Q > K → reverse; reactants increase • Decrease pressure (gas): shift to side with more moles of gas • Increase temperature: shift to endothermic side (changes K) • Catalyst: no effect on K or equilibrium position; reaches equilibrium faster
✏️

Exercises

🧠

Quiz — Unit 8: Equilibrium

Unit 8: Equilibrium

25 Qs
Q1

The equilibrium constant KἾ depends on:

KἾ depends only on temperature. Changing concentration or pressure shifts position but not K. Catalyst has no effect on K.
Q2

For 2A ⇌ B, KἾ = 25. The KἾ for B ⇌ 2A is:

Reversing equation: KἾ’ = 1/KἾ = 1/25 = 0.04.
Q3

Pure solids and liquids are omitted from K expressions because:

Activity of pure solid/liquid = 1 (standard state). Including them would multiply K by 1 with no effect; convention omits them. Only dissolved species and gases have variable activity.
Q4

If Q > K, the reaction will:

Q > K: too many products relative to equilibrium → reverse reaction reduces [products] until Q = K.
Q5

The relationship Kρ = KἾ(RT)∆ⁿ applies when:

Kρ = KἾ(RT)∆ⁿ. Δn = gaseous product moles − gaseous reactant moles. If Δn=0, Kρ=KἾ; if Δn>0, Kρ>KἾ (units of pressure built in).
Q6

For an exothermic reaction at equilibrium, increasing temperature:

Exothermic: heat is a product. Adding heat → shift left (Le Chatelier) → K decreases (fewer products at equilibrium at higher T). van’t Hoff equation: dlnK/dT = ΔH°/RT²; negative ΔH° → dlnK/dT < 0.
Q7

The ICE method stands for:

ICE table: Initial concentrations, Change (expressed as ±x), Equilibrium ([reactant]₂ = [reactant]ᵠ − x, etc.). Used to solve KἾ problems.
Q8

H₂(g) + I₂(g) ⇌ 2HI(g). KἾ = 49. If [H₂]=[I₂]=0.10 mol/L initially, [HI] at equilibrium is approximately:

K = [HI]²/(x²) where x = [H₂]ᵃ. Let γ=fraction reacted. [HI]=2×0.10×γ; [H₂]=[I₂]=0.10(1-γ). K = (0.20γ)²/(0.10(1-γ))²=49. 0.20γ/0.10(1-γ)=7. 2γ=7−7γ; 9γ=7; γ=7/9=0.778. [HI]=0.20×0.778=0.156≈0.14? Let me recalc: 2(0.10)(7/9) = 0.1556. Closest: 0.14 (a, approx).
Q9

Adding a catalyst to an equilibrium mixture:

Catalyst lowers Ea for both forward and reverse equally → equilibrium reached faster → K unchanged → position unchanged.
Q10

For PCl⁵ ⇌ PCl₃ + Cl₂, KἾ = Kρ when:

Kρ = KἾ(RT)∆ⁿ. Kρ = KἾ only when Δn = 0 (RT)⁰ = 1. For this reaction Δn=+1, so Kρ ≠ KἾ.
Q11

Which statement about Le Chatelier's principle is correct?

Le Chatelier: system at equilibrium partially opposes changes (composition, pressure, T) to restore equilibrium. Qualitative principle only — doesn’t give rate or magnitude of shift.
Q12

The KἾ expression for N₂(g) + O₂(g) ⇌ 2NO(g) is:

KἾ = [NO]² / ([N₂][O₂]). Products over reactants, each raised to stoichiometric coefficient.
Q13

At 500 K, KἾ=0.040 for a reaction. At 700 K, KἾ=0.80. The reaction is:

K increases with T (0.040→0.80 as T rises 500→700 K) → endothermic reaction (higher T favours products for endothermic). van’t Hoff: dlnK/dT = ΔH°/RT² > 0 → ΔH° > 0.
Q14

Removing a product from an equilibrium mixture causes:

Removing product: [product] decreases → Q < K → forward reaction until Q = K again. This is used industrially (e.g. removing NH₃ in Haber process, removing H₂O in esterification).
Q15

KἾ for A + B ⇌ C is 10. KἾ for 2A + 2B ⇌ 2C is:

Multiplying equation by 2: KἾ’ = (KἾ)² = 10² = 100.
Q16

The partial pressure of a gas in a mixture is:

pᵧ = yᵧ × P(total) where yᵧ = mole fraction of gas i. Dalton’s law.
Q17

For the equilibrium CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), Δn is:

Δn = (1+1) − (1+3) = 2 − 4 = −2.
Q18

KἾ = 1.0×10⁻³² means:

K << 1 (10⁻³² ≈ 0) → virtually no products; equilibrium strongly favours reactants. The reaction essentially does not occur under standard conditions.
Q19

Increasing pressure on PCl⁵ ⇌ PCl₃ + Cl₂ will:

Reaction has Δn=+1 (1 mol gas → 2 mol gas). Increasing P: Le Chatelier shifts to side with fewer moles of gas → left → less PCl⁵ decomposes.
Q20

The expression KἾ = [NH₃]² / ([N₂][H₂]³) corresponds to the equation:

N₂ + 3H₂ ⇌ 2NH₃: KἾ = [NH₃]²/([N₂][H₂]³). Match coefficients to exponents.
Q21

The Haber process operates at 200 atm because:

4 mol gas → 2 mol gas. High P shifts to fewer moles side (Le Chatelier) → higher KἾ effective yield. Beyond 200 atm, cost/safety outweighs further yield gains.
Q22

For 2NO₂(g) ⇌ N₂O₄(g), Kρ = 8.8 atm⁻¹ at 25°C. Increasing [NO₂] will:

Adding NO₂: [NO₂] increases → Q = [N₂O₄]/[NO₂]² decreases below K → Q < K → forward reaction → more N₂O₄ forms until Q=K.
Q23

A catalyst is valuable in industrial equilibrium processes because:

Catalyst does not change K or position. Value: equilibrium reached faster at lower T. Lower T means lower energy cost. Without catalyst, would need impractically high T or very long reaction time.
Q24

Which technique is used to determine whether a gaseous system has reached equilibrium?

Equilibrium: no further change in concentrations (or observable properties like colour intensity, pressure, density) over time at constant T. Dynamic equilibrium — reactions still occur but at equal rates.

Unit 8 Quiz — Chemical Equilibrium (25 Questions)

Select one answer each
Q1

A dynamic equilibrium is reached when:

At equilibrium, rate_forward = rate_reverse. Concentrations are constant (not zero). System is dynamic, not static.
Q2

The equilibrium constant Kc for aA + bB ⇌ cC + dD is:

Kc = products over reactants, each raised to their stoichiometric coefficient. Products in numerator, reactants in denominator.
Q3

If Kc >> 1, the equilibrium position lies:

Large Kc: [products] >> [reactants] at equilibrium. Very small Kc: mostly reactants. Kc = 1: roughly equal amounts.
Q4

Le Chatelier's Principle states that if a system at equilibrium is disturbed:

Adding reactant → shifts right. Removing product → shifts right. Increasing pressure (fewer gas moles side) → shifts that way.
Q5

Increasing temperature shifts an endothermic equilibrium:

Endothermic: heat is a 'reactant'. Adding heat (↑T) shifts equilibrium right. Kc increases for endothermic reactions with rising T.
Q6

A catalyst affects equilibrium by:

Catalyst lowers Ea for both directions equally. Position of equilibrium (Kc) unchanged — just reached more quickly.
Q7

Kp is related to Kc by:

Partial pressures and concentrations are related by p = cRT. Kp = Kc(RT)^Δn. If Δn = 0, Kp = Kc.
Q8

In the Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH = –92 kJ/mol), high pressure favours:

4 mol gas → 2 mol gas. Increasing pressure shifts equilibrium toward fewer moles (right) → more NH₃. Le Chatelier.
Q9

The Haber process uses 450°C as a compromise because:

Lower T: higher equilibrium yield (reaction exothermic) but too slow (not economic). 450°C gives acceptable rate with ~15% yield.
Q10

Adding an inert gas to a gaseous equilibrium at constant volume:

At constant V, adding inert gas raises total pressure but doesn't change partial pressures of reactants/products → no shift.
Q11

The reaction quotient Q is used to predict:

Q uses actual concentrations (not equilibrium). Q < K: more products form (shift right). Q > K: more reactants form (shift left).
Q12

Removing a product from an equilibrium mixture causes:

Removing product decreases [product] → Q < Kc → shift right. Used industrially to drive reactions to completion.
Q13

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at equilibrium, increasing [O₂] causes:

Adding O₂ → Q < Kc → forward reaction speeds up until new equilibrium. [SO₃] increases; [SO₂] decreases.
Q14

The Contact Process for H₂SO₄ uses V₂O₅ catalyst and 450°C because:

2SO₂ + O₂ ⇌ 2SO₃ is exothermic. High T: less favourable equilibrium. V₂O₅ catalyst allows lower T with good rate.
Q15

For a reaction with Δn = 0 (equal moles gas each side), Kp:

Kp = Kc(RT)^Δn. When Δn = 0: Kp = Kc × (RT)⁰ = Kc × 1 = Kc.
Q16

Solubility equilibrium: for a sparingly soluble salt, Ksp = Kc because:

Pure solids and pure liquids have activities = 1 — not included in Kc or Ksp. Ksp = [ions] raised to stoichiometric powers.
Q17

Which statement is true about a system at chemical equilibrium?

Equilibrium: macroscopically static, microscopically dynamic. Forward and reverse reactions continue at equal rates.
Q18

Van't Hoff equation: d(lnK)/dT = ΔH°/RT². This means:

If ΔH° > 0 (endothermic): d(lnK)/dT > 0 → K increases with T. Consistent with Le Chatelier (↑T favours endothermic direction).
Q19

Phase equilibrium between liquid and vapour is reached when:

In a closed system, equilibrium vapour pressure is constant at fixed T. This is the equilibrium between liquid and gas phases.
Q20

Dissociation of N₂O₄(g) ⇌ 2NO₂(g) at equilibrium: increasing pressure shifts equilibrium to:

Increasing pressure → system shifts to side with fewer moles (left, N₂O₄). [NO₂] decreases, [N₂O₄] increases.
Q21

The standard Gibbs energy change ΔG° is related to Kc by:

ΔG° = –RTlnK. If K > 1: lnK > 0 → ΔG° < 0 → spontaneous. Equilibrium lies toward products. ΔG° = 0 when K = 1.
Q22

Heterogeneous equilibria exclude pure solids and liquids from Kc because:

Pure solid/liquid activity = 1 (by convention). Including them would change the numerical value of K trivially. Standard practice is to exclude.
Q23

The equilibrium constant for the reverse reaction is:

If A ⇌ B has K = x, then B ⇌ A has K = 1/x. Reversing the equation inverts Kc.
Q24

Buffer solutions maintain nearly constant pH because they contain:

Added H⁺ reacts with A⁻; added OH⁻ reacts with HA. Large [HA] and [A⁻] reserves resist pH change.
Q25

Temperature is the only factor that changes the value of Kc because:

Kc depends only on T (through ΔG° = ΔH° – TΔS°). Adding reactants or changing pressure changes Q but not K.
📝 Go to Unit Test →
📝

Unit Test — 50 marks

Section A — Short Answer

30 marks
Q1 [5 marks]

Write KἾ and Kρ for: (a) 2SO₂(g)+O₂(g)⇌2SO₃(g) [2]; (b) CaCO₃(s)⇌CaO(s)+CO₂(g) [1]; Calculate Kρ from KἾ=500 at 750 K for (a). [2]

(a) KἾ=[SO₃]²/([SO₂]²[O₂]); Kρ=p(SO₃)²/(p(SO₂)²p(O₂)). Δn=2−3=−1. Kρ=KἾ(RT)⁻¹=500/(0.0821×750)=500/61.6=8.12 atm⁻¹. (b) KἾ=[CO₂]; Kρ=p(CO₂). (Solids omitted.)
Q2 [5 marks]

At 430°C, KἾ=54 for H₂(g)+I₂(g)⇌2HI(g). Initial: [H₂]=0.240, [I₂]=0.240, [HI]=0. Find [HI] at equilibrium. [5]

Let x = mol/L of H₂ that reacts. ICE: [H₂]=0.240−x, [I₂]=0.240−x, [HI]=2x. K=(2x)²/((0.240−x)²)=54. Take sqrt: 2x/(0.240−x)=√54=7.35. 2x=7.35(0.240−x)=1.764−7.35x. 9.35x=1.764. x=0.1887. [HI]=2×0.1887=0.377 mol/L.
Q3 [5 marks]

State Le Chatelier's principle. Apply it to the equilibrium: CO(g)+H₂O(g)⇌CO₂(g)+H₂(g), ΔH=−41 kJ/mol. State the effect on yield of CO of: (a) increasing T; (b) increasing P; (c) adding H₂O; (d) removing CO₂; (e) adding a catalyst.

Le Chatelier: a system at equilibrium responds to a disturbance by partially opposing it. CO reaction: Δn=0 (equal moles both sides). (a) Increase T: shifts endothermic direction (reverse); CO₂,H₂ → CO,H₂O. Yield of CO increases, K decreases. (b) Increase P: Δn=0 → no effect on position. (c) Add H₂O: Q<K → forward → more CO consumed, CO₂ and H₂ produced → yield of CO decreases. (d) Remove CO₂: Q<K → forward → more CO consumed → CO yield decreases. (e) Catalyst: no effect on equilibrium position or yield; equilibrium reached faster.
Q4 [5 marks]

Distinguish between KἾ and Q. If KἾ=0.020 for A⇌2B and a mixture has [A]=0.50, [B]=0.10, determine the direction of reaction and estimate the equilibrium [B]. [5]

KἾ uses equilibrium concentrations; Q uses current concentrations. Q=[B]²/[A]=(0.10)²/0.50=0.010/0.50=0.020. Q=0.020=K → system already at equilibrium! No net reaction needed. [B]ᵃ = 0.10 mol/L (it is already at equilibrium).
Q5 [10 marks]

The Haber process: N₂(g)+3H₂(g)⇌2NH₃(g), ΔH=−92 kJ/mol. (a) Write KἾ and Kρ, stating the relationship between them. [3] (b) Explain the compromise industrial conditions (T=400–500°C, P=200 atm, Fe catalyst). [3] (c) KἾ=0.159 at 500°C. Calculate [NH₃] at equilibrium if [N₂]=1.50 and [H₂]=3.50 mol/L. [4]

(a) KἾ=[NH₃]²/([N₂][H₂]³); Kρ=p(NH₃)²/(p(N₂)·p(H₂)³). Relationship: Kρ=KἾ(RT)∆ⁿ; Δn=2−4=−2. Kρ=KἾ(RT)⁻² → Kρ<KἾ at room T. (b) Temperature: 400–500°C is a compromise. Exothermic → lower T → higher K → better equilibrium yield; but too slow without catalyst. Fe catalyst makes rate acceptable at 400–500°C. Pressure: 200 atm increases yield (4→2 mol gas, Le Chatelier) and rate (collision frequency); above 200 atm: construction cost and safety excessive, marginal yield gain. Catalyst: Fe (promoted by K₂O, Al₂O₃) lowers Ea, doesn't change K but allows acceptable rate at lower T. (c) K=[NH₃]²/([N₂][H₂]³)=0.159. [NH₃]²=0.159×1.50×(3.50)³=0.159×1.50×42.875=0.159×64.3=10.22. [NH₃]=√10.22=3.20 mol/L.

Section B — Extended Response

20 marks
Q6 [10 marks]

(a) Derive the expression for KἾ from the law of mass action for the general reaction aA+bB⇌cC+dD, explaining the basis of the law. [3] (b) Explain, using the van’t Hoff equation d(lnK)/dT = ΔH°/RT², how K changes with temperature for exo and endothermic reactions. [3] (c) KἾ=1.00×10⁻³² for N₂(g)+O₂(g)⇌2NO(g) at 25°C but K=0.050 at 2000°C. Explain this difference and its implications for atmospheric chemistry. [4]

(a) Law of mass action (Guldberg & Waage, 1864): rate of reaction proportional to product of active concentrations of reactants each raised to stoichiometric power. At equilibrium: rate(forward) = rate(reverse). rateẟ = kẟ[A]ἐ[B]ἡ; rateᵣ = kᵣ[C]ἠ[D]ἡ. Setting equal: kẟ[A]ἐ[B]ἡ = kᵣ[C]ἠ[D]ἡ → K = kẟ/kᵣ = [C]ἠ[D]ἡ/[A]ἐ[B]ἡ. K is the ratio of rate constants, hence temperature-dependent only. (b) van’t Hoff: d(lnK)/dT = ΔH°/(RT²). Exothermic (ΔH°<0): d(lnK)/dT < 0 → K decreases as T increases → equilibrium less product-favoured at higher T. Endothermic (ΔH°>0): d(lnK)/dT > 0 → K increases with T → more products at higher T. Integrated: ln(K₂/K₁) = -(ΔH°/R)(1/T₂ - 1/T₁). (c) At 25°C: K=10⁻³² → essentially no NO at room temperature (N₂+O₂ do not react). At 2000°C (lightning, car engines): K=0.050 → measurable NO produced. Endothermic reaction → favoured at high T. Implications: lightning produces NO → contributes to natural nitrogen cycle. Car engines (combustion ~2000°C) produce NO → atmospheric chemistry: NO+½O₂→NO₂ (nitrogen dioxide, brown gas); NO₂+H₂O→HNO₃ (acid rain). Car catalytic converters reduce NO back to N₂: 2NO+2CO→N₂+2CO₂.
Q7 [10 marks]

The decomposition of N₂O₄: N₂O₄(g) ⇌ 2NO₂(g) is studied at 25°C. (a) Write KἾ and Kρ and state the relationship between them. [2] (b) At equilibrium in a 1.0 L flask: 0.10 mol N₂O₄ and 0.040 mol NO₂. Calculate KἾ. [2] (c) If an extra 0.050 mol N₂O₄ is added to the flask, calculate Q and predict the direction of reaction. [3] (d) Calculate the new equilibrium concentrations after the addition. [3]

(a) KἾ=[NO₂]²/[N₂O₄]; Kρ=p(NO₂)²/p(N₂O₄). Δn=2−1=+1. Kρ=KἾ(RT)¹=KἾ×0.0821×298=24.5KἾ. (b) KἾ=(0.040)²/0.10=0.0016/0.10=0.0160. (c) After addition: [N₂O₄]=(0.10+0.050)/1.0=0.150 M; [NO₂] still 0.040 M. Q=[NO₂]²/[N₂O₄]=(0.040)²/0.150=0.0016/0.150=0.0107. Q=0.0107 < K=0.0160 → forward reaction (N₂O₄ decomposes to make more NO₂). (d) Let x mol/L of N₂O₄ react: [N₂O₄]ᵃ=0.150−x; [NO₂]ᵃ=0.040+2x. K=0.0160=(0.040+2x)²/(0.150−x). Approx (x small): (0.040+2x)²≈0.0160×0.150=0.0024. 0.040+2x=√0.0024=0.0490. 2x=0.0090; x=0.0045. [N₂O₄]ᵃ=0.145 M; [NO₂]ᵃ=0.0490 M. Check: (0.049)²/0.145=0.00240/0.145=0.0166 ≈ 0.0160 ✓.
← Unit 7: Solvent & ColligativeUnit 9: pH →

💳 Pay to Download

Unit notes — instant PDF download

Price
300 RWF
One-time · Download immediately after payment
Phone Number
Rwanda — MTN MoMo or Airtel Money
Network
MTN MoMo
Airtel Money
Other
Processing Payment...
Check your phone — enter your PIN to confirm.
✅
Payment Confirmed!
Generating your PDF...
❌
Payment Not Completed
WhatsApp
⚙️
Payment service is being set up.
Online payment is not yet available for this unit.
Please contact your teacher to get access.